1. Identify the half-reactions involved

The given cell reaction is: $Fe^{2+}(aq) + Ag^+(aq) \rightarrow Fe^{3+}(aq) + Ag(s)$

So:

• $Ag^+$ is reduced to $Ag$
• $Fe^{2+}$ is oxidized to $Fe^{3+}$

Hence, $E^\circ_{cell} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode (reduction form)}}$

Thus, $E^\circ_{cell} = E^\circ_{Ag^+/Ag} - E^\circ_{Fe^{3+}/Fe^{2+}}$

2. Use the given standard reduction potentials

Given: $E^\circ_{Ag^+/Ag} = x$ $E^\circ_{Fe^{2+}/Fe} = y$ $E^\circ_{Fe^{3+}/Fe} = z$

We need $E^\circ_{Fe^{3+}/Fe^{2+}}$.

3. Relate $E^\circ_{Fe^{3+}/Fe^{2+}}$ with the given values

Consider the two reductions:

$Fe^{3+} + 3e^- \rightarrow Fe \qquad E^\circ = z$ $Fe^{2+} + 2e^- \rightarrow Fe \qquad E^\circ = y$

Also, $Fe^{3+} + e^- \rightarrow Fe^{2+} \qquad E^\circ = ?$

Using standard Gibbs energy relation: $\Delta G^\circ = -nFE^\circ$

For the 3-electron reduction: $\Delta G_1^\circ = -3Fz$

For the 2-electron reduction: $\Delta G_2^\circ = -2Fy$

And since $Fe^{3+} + 3e^- \rightarrow Fe$ is the sum of $Fe^{3+} + e^- \rightarrow Fe^{2+}$ and $Fe^{2+} + 2e^- \rightarrow Fe$ we have $\Delta G_1^\circ = \Delta G^\circ_{Fe^{3+}/Fe^{2+}} + \Delta G_2^\circ$

So, $-3Fz = \Delta G^\circ_{Fe^{3+}/Fe^{2+}} - 2Fy$

Therefore, $\Delta G^\circ_{Fe^{3+}/Fe^{2+}} = -3Fz + 2Fy = -F(3z-2y)$

Hence, $E^\circ_{Fe^{3+}/Fe^{2+}} = 3z - 2y$

4. Calculate the cell potential

Now, $E^\circ_{cell} = x - (3z - 2y)$ $E^\circ_{cell} = x + 2y - 3z$

5. Match with the options

$\boxed{E^\circ_{cell} = x + 2y - 3z}$

So the correct option is: $\boxed{A}$